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Forums - General Discussion - Does 0.999... Equal 1?

Yes it does.



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vlad321 said:

So here it goes, I hope this is not a joke thread because I'm about to waste 3 minutes of my life writing this up. From here on out S is sigma

0.999999 can be represented as the infinite sum of S(9 * (1/10) ^n) where n = 1 to infinity.This is basically sayying that 0.99999... = 9/10 +9/100 +9/1000 +....

Meanwhile the sum of a geometric series is PROVEN to be equal to :

a/(1-(1/r))

Where a is the first term, 0.9 in this case, and r is the rate of growth, 1/10 in this case. So where does that leave us?

0.9/(1-(1/10) =

= 0.9/( 9/10) =

= 0.9/0.9 =

= 1

 

 

And that is the easiest and way to prove this. The whole 0.3333 = 1/3rd is crap and not a real proof by any stretch of the imagination.

Of course that it is proof. You can't belittle it just because it doesn't have a pretty geometric equation to it. Sure, what you wrote down is true, but you can't do, is prove that 0.(3)= 1/3 isn't true. Why? You would do in the whole 0.(9) theory, and that you can't do, since you proved it.

Math is not about making things complicated, sheesh..

 



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Any day now, the dollar will be worth less than 2 zloty......any day now.....and my life savings will be in total jepordy ;(.

Rath said:
steven787 said:
HappySqurriel said:
 

It really isn't ...

On the left hand side of my equation I will deal with numbers using fractions, on the right hand side of my equation I will deal with numbers using decimals

1 = 1

1/3 = 0.333...

(1/3)*3 = (0.333...)*3

1 = 0.999...

 

 

 

Now, I see that you guys want to keep insisting that 1/3 = .333333.  it doesn't it's ~.

 

I would argue that 0.(3) is the decimal representation of 1/3 as it is infinitely long. Basically it's better to present a recurring number as a fraction as it's less ambigious but they are equivalent.

 

The truth is that fraction and decimal representation of numbers both have their flaws and that (most) mathematicians are very aware of that ... The problem (at the moment) is there are no other representations which are better



vlad321 said:

So here it goes, I hope this is not a joke thread because I'm about to waste 3 minutes of my life writing this up. From here on out S is sigma

0.999999 can be represented as the infinite sum of S(9 * (1/10) ^n) where n = 1 to infinity.This is basically sayying that 0.99999... = 9/10 +9/100 +9/1000 +....

Meanwhile the sum of a geometric series is PROVEN to be equal to :

a/(1-(1/r))

Where a is the first term, 0.9 in this case, and r is the rate of growth, 1/10 in this case. So where does that leave us?

0.9/(1-(1/10) =

= 0.9/( 9/10) =

= 0.9/0.9 =

= 1

 

 

And that is the easiest and way to prove this. The whole 0.3333 = 1/3rd is crap and not a real proof by any stretch of the imagination.

"n" is a theoretical number, so the proof is invalid in every type of math not relating to geometry as perceived by people.



I would cite regulation, but I know you will simply ignore it.

steven787 said:
vlad321 said:

So here it goes, I hope this is not a joke thread because I'm about to waste 3 minutes of my life writing this up. From here on out S is sigma

0.999999 can be represented as the infinite sum of S(9 * (1/10) ^n) where n = 1 to infinity.This is basically sayying that 0.99999... = 9/10 +9/100 +9/1000 +....

Meanwhile the sum of a geometric series is PROVEN to be equal to :

a/(1-(1/r))

Where a is the first term, 0.9 in this case, and r is the rate of growth, 1/10 in this case. So where does that leave us?

0.9/(1-(1/10) =

= 0.9/( 9/10) =

= 0.9/0.9 =

= 1

 

 

And that is the easiest and way to prove this. The whole 0.3333 = 1/3rd is crap and not a real proof by any stretch of the imagination.

"n" is a theoretical number, so the proof is invalid.

 

No it's not, n is just what you are multiplying 1/10 by each step of they way until inifnity. It's not theoretical in any ways, it's part of the Sigma Notation. Fail more.



Tag(thx fkusumot) - "Yet again I completely fail to see your point..."

HD vs Wii, PC vs HD: http://www.vgchartz.com/forum/thread.php?id=93374

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gamrReview's broken review scores: http://gamrconnect.vgchartz.com/post.php?id=4170835

 

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to the n, as in to infinity or to the theoretical non-end of numbers.



I would cite regulation, but I know you will simply ignore it.

steven787 said:
vlad321 said:

So here it goes, I hope this is not a joke thread because I'm about to waste 3 minutes of my life writing this up. From here on out S is sigma

0.999999 can be represented as the infinite sum of S(9 * (1/10) ^n) where n = 1 to infinity.This is basically sayying that 0.99999... = 9/10 +9/100 +9/1000 +....

Meanwhile the sum of a geometric series is PROVEN to be equal to :

a/(1-(1/r))

Where a is the first term, 0.9 in this case, and r is the rate of growth, 1/10 in this case. So where does that leave us?

0.9/(1-(1/10) =

= 0.9/( 9/10) =

= 0.9/0.9 =

= 1

 

 

And that is the easiest and way to prove this. The whole 0.3333 = 1/3rd is crap and not a real proof by any stretch of the imagination.

"n" is a theoretical number, so the proof is invalid.

It's a theoretical number, but one that exists in reality, that belongs to the so called "reality numbers".It can range from one to any postive number known that is not a fracture,or anything as strange (don't have the vocab to say what i want, but you understand i hope). His equation is valid.

 



Huh. Who would've thought that beggining anew in my real life would coincide with starting anew on vgchartz?

Any day now, the dollar will be worth less than 2 zloty......any day now.....and my life savings will be in total jepordy ;(.

Menago KF said:
steven787 said:
vlad321 said:

So here it goes, I hope this is not a joke thread because I'm about to waste 3 minutes of my life writing this up. From here on out S is sigma

0.999999 can be represented as the infinite sum of S(9 * (1/10) ^n) where n = 1 to infinity.This is basically sayying that 0.99999... = 9/10 +9/100 +9/1000 +....

Meanwhile the sum of a geometric series is PROVEN to be equal to :

a/(1-(1/r))

Where a is the first term, 0.9 in this case, and r is the rate of growth, 1/10 in this case. So where does that leave us?

0.9/(1-(1/10) =

= 0.9/( 9/10) =

= 0.9/0.9 =

= 1

 

 

And that is the easiest and way to prove this. The whole 0.3333 = 1/3rd is crap and not a real proof by any stretch of the imagination.

"n" is a theoretical number, so the proof is invalid.

It's a theoretical number, but one that exists in reality, that belongs to the so called "reality numbers".It can range from one to any postive number known that is not a fracture,or anything as strange (don't have the vocab to say what i want, but you understand i hope). His equation is valid.

 

You mean complex, as in it includes sqrt(-1)?

Anyways, there are a lot more theoretical things than will blow your mind than a simple sigma notation statement. Did you know that some infinities are greater than others? While the real numbers are infinite, the numbers which are not whole are more numerous than the reals?

The question of infinity was put to rest when Newton came along with his epsilon-delta definition.

 



Tag(thx fkusumot) - "Yet again I completely fail to see your point..."

HD vs Wii, PC vs HD: http://www.vgchartz.com/forum/thread.php?id=93374

Why Regenerating Health is a crap game mechanic: http://gamrconnect.vgchartz.com/post.php?id=3986420

gamrReview's broken review scores: http://gamrconnect.vgchartz.com/post.php?id=4170835

 

Menago, I understand.

It really comes down to this.

If we want to use the representation of .(3) = 1/3, then
.(3) + .(3) + .(3) = 1 and not .(9)

It can't be both. In the real world a number can't have two values.

We have to pick the set of rules ahead of time (in this case: .(3) = 1/3), and justify the rules with other rules (in this case, the value of n).



I would cite regulation, but I know you will simply ignore it.

steven787 said:
Menago, I understand.

It really comes down to this.

If we want to use the representation of .(3) = 1/3, then
.(3) + .(3) + .(3) = 1 and not .(9)

It can't be both. In the real world a number can't have two values.

We have to pick the set of rules ahead of time (in this case: .(3) = 1/3), and justify the rules with other rules (in this case, the value of n).

 

Wait, what's wrong?



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